Question

he models the volume, v(x), in cubic feet, as a function of the length x, in feet v(x) = 2x(x - 1)² he created the diagram and graph to model the box and its volume. the zeros of function v occur at 0 and 1. to make the box, samson should use a value for x: either between 0 and 1 or greater than 1, between 0 and 1, greater than 1

Explanation:

Step1: Find zeros of \( v(x) \)

To find the zeros of \( v(x) = 2x(x - 1)^2 \), set \( v(x)=0 \).
\( 2x(x - 1)^2 = 0 \) implies \( 2x = 0 \) or \( (x - 1)^2 = 0 \).
Solving \( 2x = 0 \) gives \( x = 0 \). Solving \( (x - 1)^2 = 0 \) gives \( x = 1 \) (double root). So zeros are at \( 0 \) and \( 1 \).

Step2: Analyze volume for box

Volume \( v(x) \) must be positive (since length can't be negative and volume of a box is positive).
- For \( x \) between \( 0 \) and \( 1 \): Let \( x = 0.5 \), \( v(0.5)=2(0.5)(0.5 - 1)^2 = 1(0.25)=0.25>0 \).
- For \( x > 1 \): Let \( x = 2 \), \( v(2)=2(2)(2 - 1)^2 = 4(1)=4>0 \).
- For \( x < 0 \): Length can't be negative, so \( x \) must be positive.

But the box’s dimensions (from the diagram, like \( 2 \) ft, \( 4 \) ft) suggest \( x \) should make sense for a physical box. However, mathematically, \( v(x)>0 \) when \( x > 0 \) and \( x eq 1 \), but since \( x = 0 \) or \( x = 1 \) give volume \( 0 \) (not a box), \( x \) must be either between \( 0 \) and \( 1 \) or greater than \( 1 \) (but checking the graph, the volume is positive in those intervals). Wait, but the first part’s zero is \( 0 \) and \( 1 \). For the second part, to make a box, volume must be positive, so \( x \) should be either between \( 0 \) and \( 1 \) or greater than \( 1 \)? Wait, no—wait the graph: when \( x > 1 \), the graph is below? Wait no, the function \( v(x)=2x(x - 1)^2 \): when \( x > 1 \), \( x \) is positive, \( (x - 1)^2 \) is positive, so \( v(x) \) is positive. Wait the graph in the image: maybe I misread. Wait the graph has \( v(x) \) negative? No, volume can't be negative. Wait maybe the function is \( v(x)= -2x(x - 1)^2 \)? Wait no, the problem says \( v(x)=2x(x - 1)^2 \). Wait maybe the diagram: the first part’s zero is \( 0 \) and \( 1 \). Then, to make a box, \( x \) must be such that volume is positive. So \( x > 0 \) (since length can't be negative) and \( x eq 1 \). But the options are "between 0 and 1", "greater than 1", or "either between 0 and 1 or greater than 1". Wait, let's re-express \( v(x)=2x(x - 1)^2 \). The square term is always non-negative, \( 2x \) is positive when \( x > 0 \). So \( v(x) \geq 0 \) for \( x \geq 0 \), with zeros at \( x = 0 \) and \( x = 1 \). So for \( x > 0 \), \( x eq 1 \), \( v(x) > 0 \). So \( x \) can be between \( 0 \) and \( 1 \) (since \( x > 0 \)) or greater than \( 1 \). But maybe the diagram’s dimensions (like the original sheet) have constraints. Wait the first blank: zeros at \( 0 \) and \( 1 \). The second blank: either between \( 0 \) and \( 1 \) or greater than \( 1 \)? Wait no, maybe I made a mistake. Wait the graph: if the graph is as shown (red curve), when \( x > 1 \), is \( v(x) \) positive? Wait the function \( v(x)=2x(x - 1)^2 \) is a cubic with a double root at \( x = 1 \) and a root at \( x = 0 \). The leading term is \( 2x^3 \), so as \( x \to \infty \), \( v(x) \to \infty \), and as \( x \to 0^+ \), \( v(x) \to 0^+ \). So the graph should rise from \( 0 \) at \( x = 0 \), touch \( x = 1 \) (double root), then rise. But the given graph looks like it’s decreasing after \( x = 1 \). Wait maybe the function is \( v(x)= -2x(x - 1)^2 \)? Then zeros at \( 0 \) and \( 1 \), and for \( x \) between \( 0 \) and \( 1 \), \( v(x) > 0 \) (since \( -2x \) is negative, \( (x - 1)^2 \) is positive, so negative times positive is negative? No, that can't be. Wait the problem says "volume", so it must be positive. So maybe the function is \( v(x)=2x(1 - x)^2 \)? Then zeros at \( 0 \) and \( 1 \), and for \( x \) between \( 0 \) and \( 1 \), \( v(x) > 0 \), and for \( x > 1 \), \( v(x) < 0 \) (since \( 1 - x \) is negative, squared is positive, \( 2x \) positive, so positive times positive is positive? No, \( (1 - x)^2 \) is positive, \( 2x \) positive, so \( v(x) \) positive. I'm confused. Wait the first part: zeros at \( 0 \) and \( 1 \). The second part: to make a box, volume must be positive, so \( x \) can't be \( 0 \) or \( 1 \), so \( x \) is either between \( 0 \) and \( 1 \) or greater than \( 1 \)? But the options include "either between 0 and 1 or greater than 1". Wait the first blank: zeros at \( 0 \) and \( 1 \). The second blank: either between \( 0 \) and \( 1 \) or greater than \( 1 \).

Answer:

Zeros at \( 0 \) and \( 1 \); \( x \) should be either between \( 0 \) and \( 1 \) or greater than \( 1 \) (but following the first part’s zero: \( 0 \) and \( 1 \); second part: either between \( 0 \) and \( 1 \) or greater than \( 1 \)).

(For the first blank: \( 0 \) and \( 1 \); second blank: "either between 0 and 1 or greater than 1")